Download 18.781 Homework 2

18.781 Homework 2
Due: 18th February 2014
Q 1 (1.3(3)). Prove that a number is divisible by 2 if and only if the units digit is divisible by 2; that a
number is divisible by 4 if and only if the integer formed by its tens and units digit is divisible by 4; that a
number is divisible by 8 if and only if the the integer formed by the last three digits is divisible by 8.
Pk
Proof. Let the decimal expansion of a number n be i=0 ai 10i = a0 + 10a1 + 100a2 + . . . + 10k ak . Note
Pk
P
P
k
k
i
i
i
that 10 |
i=1 ai 10 , 100 |
i=2 ai 10 , 1000 |
i=3 ai 10 . These can be rewritten as n = a0 + 10k =
a0 + 10a1 + 100l = a0 + 10a1 + 100a2 + 1000m for some integers k, l and m. As 2 | 10k, 2 | a0 if and only if
2 | n. As 4 | 100l, 4 | n if and only if 4 | a0 + 10a1 . As 8 | 1000m, 8 | n if and only if 8 | a0 + 10a1 + 100a2 .
Q 2 (1.3(10)). Prove that any positive integer of the form 3k + 2 has a prime factor of the same form;
similarly for each of the forms 4k + 3 and 6k + 5.
Proof. Every number can be written as 3q + r for r ∈ {0, 1, 2} by the division algorithm. Any prime factor
of a number of the form 3k + 2 which is not of the same form has to be of the form 3l + 1 as 3 - 3k + 2 for
any k (the only prime of the form 3l is 3). A product of numbers of the form 3l + 1 is again a number of the
same form (by looking at all products modulo 3, and using Theorem 2.1(4)). So the fundamental theorem
of arithmetic along with the fact above tells us that a number of the form 3k + 2 must have a prime factor
of the form 3k + 2.
A prime factor of a number of the form 4k + 3 which is not of the same form has to be of the form 4l + 1
(2 - 4k + 3 for any k; any prime factor of a number of the form 4k + 3 has to be odd and therefore cannot be
divisible by 4 or be congruent to 2 (mod 4)) and once again a product of numbers congruent to 1 (mod 4)
is again congruent to 1 (mod 4). So the fundamental theorem of arithmetic along with this fact tell us that
a number of the form 4k + 3 must have a prime factor of the same form.
Note that both 3 and 2 do not divide a number of the form 6k + 5 = 2(3k + 2) + 1 = 3(2k + 1) + 2, so any
prime factor of this number not of the same form should be of the form 6l + 1 (and cannot be of the form
6l or 6l + 2 or 6l + 3 or 6l + 4 as a number of any of these forms is either divisible by 2 or 3). A product of
numbers of the form 6l + 1 is again a number of the same form (by looking at products modulo 6). So the
the fundamental theorem of arithmetic now tells us that a number of the form 6k + 5 must have a prime
factor of the same form.
Q 3 (1.3(17)). Twin primes are those differing by 2. Show that 5 is the only prime belonging to two such
pairs. Show also that there is a one-to-one correspondence between twin primes and numbers n such that
n2 − 1 has just four positive divisors.
Proof. 5 is clearly the smallest prime belonging to two such pairs (3, 5) and (5, 7). Let p ≥ 7 be a prime. p
has to either be of the form 3k + 1 or 3k + 2 for some k ≥ 2. If p is of the form 3k + 1, then p + 2 = 3(k + 1)
cannot be prime, and if p is of the form 3k + 2, then p − 2 = 3k cannot be prime. For p to belong two pairs
of twin primes, both p − 2 and p + 2 must be prime and this isn’t possible for a prime p ≥ 7.
Consider the pair (n − 1, n + 1). We claim that this gives us a pair of twin primes if and only if n2 − 1
has exactly four positive divisors.
Assume first that n − 1 and n + 1 are both prime. Any prime divisor of n2 − 1 = (n − 1)(n + 1) has to
divide either n − 1 or n + 1 and therefore has to equal one of them (as n − 1 and n + 1 are prime, and don’t
have any proper divisors). Any prime factor of a proper divisor d of n2 − 1 is also a prime divisor of n2 − 1,
so d = (n − 1)l or (n + 1)l. Now, if dd0 = n2 − 1, then we either have d0 l = n + 1 or d0 l = n − 1 (both n − 1
1
and n + 1 are prime and therefore not zero) but as n + 1 and n − 1 are prime, this forces l = 1 as d is a
proper divisor of n2 − 1. So the four positive divisors of n2 − 1 are 1, n + 1, n − 1, n2 − 1.
If n2 − 1 has only four positive divisors then the set of positive divisors is equal to {1, n − 1, n + 1, n2 − 1}
(n > 1 as n = 1 forces n2 − 1 = 0 and it is no longer true that n2 − 1 has exactly four positive divisors.
n > 1 implies both n − 1 and n + 1 are positive). This forces both n − 1 and n + 1 to be prime as any
divisor of either of these numbers will also be a divisor of n2 − 1. This gives us the pair of twin primes
(n − 1, n + 1).
Q 4 (1.3(26)). Prove that there are infinitely many primes of the form 4n + 3 and of the form 6n + 5.
Proof. Suppose there are only finitely many primes of the form 4n + 3, say {p1 , . . . , pk }. Let P denote their
k
∼ 3k (mod 4) = 9 2 (mod 4) = 1 (mod 4) (Theorem 2.1(4)). Then
product. Suppose k is even. Then P =
∼ 3 (mod 4), so by Problem 1.3(10) has to have a prime factor of the form 4n + 3. But pi - P + 2 for
P +2 =
all 1 ≤ i ≤ k as pi | P and pi - 2. This is a contradiction. If k is odd, then P ∼
= 3k (mod 4) = 3 (mod 4).
Then, P + 4 is a number of the form 4n + 3, so it must have a prime factor of the form 4n + 3 by Problem
1.3(10). But pi - P + 4 as pi | P and pi - 4, which is a contradiction. This shows that there must be infinitely
many primes of the form 4n + 3.
The proof for the infinitude of primes of the form 6n + 5 is along the same lines. Suppose there are only
finitely many; say {p1 , . . . , pk }. Let P denote their product as before. 52 ∼
= 1 (mod 6). If k is even, then
P +4 ∼
= 25k/2 + 4 (mod 6) = 5 (mod 6) and so P + 4 must have a prime divisor of the form 6n + 5 by
Problem 1.3(10) but none of the pi s divide P + 4. Similarly, if k is odd, then P + 6 ∼
= 5 (mod 6) so P + 6
must have a prime divisor of the form 6n + 5 by Problem 1.3(10) but none of the pi s divide P + 6 as pi | P
and pi - 6 for every i.
Q 5 (1.3(41)). Prove that an odd integer n > 1 is prime if and only if it is not expressible as a sum of three
or more consecutive positive integers.
Proof. If an odd integer n is expressible as a sum of three or more consecutive positive integers, then for
some m ≥ 1 and k ≥ 3,
n = m + (m + 1) + . . . + (m + (k − 1)) = km +
k(k − 1)
2
and cannot be prime (k and m + k−1
If k is odd, then n = k. m + k−1
2
2 are integers strictly bigger than 1).
k
If k is even, then, n = 2 .(2m + (k − 1)) and once again cannot be prime as k2 and 2m + (k − 1) are integers
strictly bigger than 1. If an odd integer n is not prime, write n = a.b for some other positive integers a and
b strictly bigger than 1. a and b must be odd. Assume a ≤ b without loss of generality. Let k = a ≥ 3 and
a−1
a+1
m = b − a−1
2 ≥ a − 2 = 2 ≥ 2. Then,
m + (m + 1) + . . . + (m + (k − 1)) = km +
k(k − 1)
k−1
= k.(m +
) = a.b = n
2
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Q 6 (1.4(10)). Let S be a set of n elements. Count the number of ordered pairs (A, B) of subsets such that
∅ ⊂ A ⊂ B ⊂ S. Let c(j, k) denote the number of such ordered pairs for which A contains j elements and
B contains k elements. Show that
X
(1 + y + xy)n =
c(j, k)xj y k
0≤j≤k≤n
What does this give for x = y = 1?
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Proof. The number of ways of picking the subset B is
ways of picking A, so in total c(j, k) = nk kj .
(1 + y(1 + x))n =
n
k
and for each way of picking B there are
n X
n
k=0
k
(y(1 + x))
k
j
many
k


k X
k
=
yk 
xj 
k
j
j=0
k=0
X nk =
y k xj
k
j
0≤j≤k≤n
X
=
c(j, k)y k xj
n X
n
0≤j≤k≤n
We applied the binomial theorem twice, once in line 1 and once in line 3. For x = y = 1, this gives
X
3n =
c(j, k)
0≤j≤k≤n
Q 7 (1.4(20)). Show that if m and n are integers with 0 ≤ m < n, then
n
X
n k−1
(−1)k
= (−1)m+1
k
m
k=m+1
Pn
Proof. We will show this by induction on m. If m = 0, then we have to show k=1 (−1)k nk = −1 (as
k−1
= k−1
= 1 for all k). This follows by applying the binomial expansion for (a + b)n with a = −1 and
m
0
Pn
k
k−1
k−1
b = 1. Let Sm = k=m+1 (−1)k nk k−1
+ m−1
= m−1
.
m . Recall
m
!
n
X
k−1
k−1
m
k n
m n
Sm + S(m−1) =
(−1)
+
+ (−1)
k
m
m−1
m m
k=m+1
!
n
X
n
k
n
m
=
(−1)k
+ (−1)m
k
m
m m
k=m+1
n
X
n
k
(−1)k
=
k
m
k=m
=0
We get the last equality by plugging in y = −1 in the equality from Problem 1.4(10) and comparing the
coefficient of xm on both sides (we are using m < n here).
Q 8 (2.1(6)). Prove that if p is a prime and a2 ∼
= b2 (mod p), then p | (a + b) or p | (a − b).
Proof.
a2 ∼
= b2 (mod p) ⇔ p | (a2 − b2 ) ⇔ p | (a − b)(a + b) ⇔ p | (a + b) or p | (a − b)
Q 9 (2.1(14)). Show that 7 | 32n+1 + 2n+2 .
Proof. 9 ∼
= 2 (mod 7).
32n+1 ∼
= 3.9n (mod 7) = 3.2n (mod 7)
So
32n+1 + 2n+2 ∼
= 3.2n + 4.2n (mod 7) = 7.2n (mod 7) = 0 (mod 7)
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Q 10 (2.1(17)). Show that 61! + 1 ∼
= 63! + 1 ∼
= 0(mod 71).
Proof. 70! + 1 ∼
= 0 (mod 71) by Wilson’s theorem. −71.71 = −5041. Now,
64.65.66.67.68.69.70 ∼
= (−7).(−6).(−5).(−4).(−3).(−2).(−1) (mod 71) = −5040 (mod 71) = 1 (mod 71)
Therefore,
63! ∼
= 70! (mod 71) = −1 (mod 71)
62.63.64.65.66.67.68.69.70 ∼
= (−9).(−8).1 (mod 71) = 72 (mod 71) = 1 (mod 71)
61! ∼
= 70! (mod 71) = −1 (mod 71)
Q 11 (2.1(38)). Prove that there are infinitely many primes of the form 4n + 1.
Proof. Suppose there were only finitely many say {p1 , p2 , . . . pk }. Let P denote their product. Consider the
number N = 4P 2 + 1. A prime divisor of this number, say l, has to be odd (as N is odd) and also be different
from any of the primes we listed. Then 2P (mod l) is a square root of −1 modulo l, so l has to be a prime
of the form 4n + 1 which is a contradiction.
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